How many Gallons of a 12% indicator solution must be mixed with a20% indecator solution to get 10 gallons of a 14% solution.

przesypkai4

przesypkai4

Answered question

2022-07-27

How many Gallons of a 12% indicator solution must be mixed with a20% indecator solution to get 10 gallons of a 14% solution.

Answer & Explanation

Dominique Ferrell

Dominique Ferrell

Beginner2022-07-28Added 18 answers

This can be solved using the following system of equations,which can be solved by a variety of methods, let me know if youwant me to solve it.
.12x+.2y=.14
x+y=10
Where x is the number of gallons of the first solution, and yis the number of gallons of the second.
Israel Hale

Israel Hale

Beginner2022-07-29Added 5 answers

Let x= the number of gallons at 20%
y= the number of gallons at12%
The total number of gallons: x+y = 10 gallons.
The total added mixture to the solution is
0.20x +.12y = .14(10)
Elminating the decimals and simplifying we have
20x + 12y = 140
Now we want to find how many gallons we need at 12%. So fromthe total number of gallons, we want to eliminate x in the equation by substituting for x=10-yin order to solve for y. Hence,
20(10-y) + 12y= 140
200 -20y +12y = 140
-8y = -60
y = 7.5 gallons
Therefore, you will need to add 7.5 gallons at 12%.Hence, if you wanted to find out how many gallons at 20% you simplysolve for x in the equation x+y=10 which yields x=2.5gallons.
Checking your answer, we simply substitute the values into theequation 20x +12y= 140.
Then 20(2.5) + 12(7.5) = 140
140 = 140 is true

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